求数列{（2n1）5ⁿ}的前n项和解：a‹n›（2n1）5ⁿS‹n›15¹+35²+55³+75⁴++（2n1）5ⁿ（1）5S‹n›15²+35³+55⁴++（2n3）5ⁿ+（2n1）5^(n+1)(2)(1)(2)（错项相减）得：4S‹n›15¹+25²+25³+25⁴++25ⁿ（2n1）5^(n+1)5+2(5²+5³+⁴++5ⁿ)（2n1）5^(n+1)5+2(5¹+5²+5³+⁴++5ⁿ)（2n1）5^(n+1)5+25(5ⁿ1)4(2n1)5^(n+1)(152)+(32)2n5^(n+1)∴S‹n›（14）{(152)+(32)2n5^(n+1)}(n2)(38)5^(n+1)+(158)
